Question

"Question 2 In the given figure sides AB and AC of ΔABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.

Class 9 - Math - Triangles Page 132"

Answer 1

Linear pair of angles:
If Non common arms of two adjacent angles form a line, then these angles are called linear pair of angles.


Axiom- 1
If a ray stands on a line, then the sum of two adjacent angles so formed is 180°i.e, the sum of the linear pair is 180°.

Axiom-2
If the sum of two adjacent angles is 180° then the two non common arms of the angles form a line.

The two axioms given above together are called the linear pair axioms.

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Use the Inequality that the side opposite to the larger angle is longer, to show the inequality.

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Given: ∆ABC is a Triangle &
∠PBC < ∠QCB

To Prove:
AC>AB


Proof:

∠ABC + ∠PBC = 180° ( by linear pair axiom)
⇒ ∠ABC = 180° – ∠PBC

also,
∠ACB + ∠QCB = 180° (by Linear Pair axiom)
⇒ ∠ACB = 180° – ∠QCB

Since,
∠PBC < ∠QCB therefore, ∠ABC > ∠ACB

Here, the side opposite to the larger angle is longer.

Hence, AC > AB

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Hope this will help you.....

Answer 2

Linear pair of angles:

If Non common arms of two adjacent angles form a line, then these angles are called linear pair of angles.

Axiom- 1

If a ray stands on a line, then the sum of two adjacent angles so formed is 180°i.e, the sum of the linear pair is 180°.

Axiom-2

If the sum of two adjacent angles is 180° then the two non common arms of the angles form a line.

The two axioms given above together are called the linear pair axioms.

____________________________________

Use the Inequality that the side opposite to the larger angle is longer, to show the inequality.

____________________________________

Given: ∆ABC is a Triangle &

∠PBC < ∠QCB

To Prove:

AC>AB

Proof:

∠ABC + ∠PBC = 180° ( by linear pair axiom)

⇒ ∠ABC = 180° – ∠PBC

also,

∠ACB + ∠QCB = 180° (by Linear Pair axiom)

⇒ ∠ACB = 180° – ∠QCB

Since,

∠PBC < ∠QCB therefore, ∠ABC > ∠ACB

Here, the side opposite to the larger angle is longer.

Hence, AC > AB